Let $\Phi$ be a (crystallographic) root system with Weyl group $\mathcal{W}$, and $\Phi^+$ a choice of positive roots, and $$ q := \prod_{\alpha\in\Phi^+} (\exp(\alpha/2) - \exp(-\alpha/2)) = \sum_{w\in\mathcal{W}} \mathrm{sgn}(w)\,\exp(w\rho) $$ be the denominator in the Weyl character formula ("WCF"); here, of course, $\exp$ is a formal exponential, $\rho := \frac{1}{2}\sum_{\alpha\in\Phi^+}\alpha = \sum_i\varpi_i$ is the Weyl vector, $\varpi_i$ are the fundamental weights, and $\mathrm{sgn}$ is the abelian character of $\mathcal{W}$ with value $-1$ on the reflections.

Since $q$ is $\mathcal{W}$-anti-invariant (we have $w(q) = \mathrm{sgn}(w)\,q$), it follows that $q^2$ is $\mathcal{W}$-invariant. So $q^2$ can be expressed as a polynomial in the fundamental characters (the fundamental characters being the $x_i := q^{-1} \sum_{w\in\mathcal{W}} \mathrm{sgn}(w)\,\exp(w(\rho+\varpi_i))$ by the WCF); or equivalently, as a polynomial in the averages of the fundamental weights (meaning the $\frac{1}{\#\mathcal{W}} \sum_{w\in\mathcal{W}} \exp(w\varpi_i)$): see the references to Bourbaki and Lorenz in this related question. This invariant quantity $q^2$ can be considered as a kind of discriminant (see PS3 below).

My question about this square denominator $q^2$ is: how can we compute it in practice (as a polynomial of the kind I just described)? Is there a convenient expression? Or perhaps, can I get LiE or Sage to compute it?

I am also interested in any sort of remarks about it: for example, does it have a standard name (beyond "the square of the Weyl denominator")? Is it the (virtual) character of some naturally defined element in the Grothendieck group of representations?

**PS 1:** I should have made it clear that I am looking for a computational approach that does not involve writing down all the $\#\mathcal{W}$ terms in $q$. (LiE is capable of doing computations on the ring of [multiplicative] $\mathcal{W}$-invariants without fully expanding them; so the question is whether we can do this for a $\mathcal{W}$-anti-invariant like $q$.)

**PS 2:** If we let $R = \mathbb{C}[\exp(\Lambda)]^{\mathcal{W}}$, where $\Lambda := \langle\Phi^\vee\rangle^*$ is the weight lattice, be the ring of multiplicative $\mathcal{W}$-invariants (which is a polynomial ring), then the ring of multiplicative $\mathcal{W}_0$-invariants, where $\mathcal{W}_0 := \ker(\mathrm{sgn})$ is the group of rotations in the Weyl group, is the free quadratic algebra $R \oplus R q$. So asking to describe $q^2$ in $R$ is the missing bit in the presentation of this algebra. Hopefully this helps motivate the question.

**PS 3:** In the case where $\Phi = A_n$, then $q^2$ expressed as a polynomial of $x_1,\ldots,x_n$ is exactly the discriminant of the polynomial (in the $z$ variable) $z^{n+1} - x_1 z^n + x_2 z^{n-1} + \cdots + (-1)^n x_n z + (-1)^{n+1}$ (indeed, consider a diagonal element of $SL_{n+1}$: then $x_1,\ldots,x_n$ give the elementary symmetric functions of the $n+1$ eigenvalues, whose product is $1$, and $q^2$ is their discriminant since $q$ is the Vandermonde determinant). What I'm asking for is a generalization of this to the other root systems.

definingfundamental characters, not exhibiting $q^2$ as such a polynomial. $\endgroup$ – LSpice Jun 2 '17 at 14:02